Question

how to find the rank of a matrix?

Answer 1

\[\left[\begin{matrix}1 & -1 &2\\ 3 & -3&6\\-2&2&4\end{matrix}\right]\]

Answer 2

even i want to learn this..
so
*bookmark*

Answer 3

Do the gauss jordan elimination method and check how many leading ones there are

Answer 4

""Suppose that A is a matrix then the row space of A and the column space of A will have the same dimension. We call this common dimension the rank of A and denote it by rank (A)."

http://tutorial.math.lamar.edu/Classes/LinAlg/FundamentalSubspaces.aspx

Answer 5

if the row and col have different dimensions, then rank (A) is equal or smaller than the smaller value

Answer 6

By inspection, it would be 2. Row 2 is a multiple of row one. row 3 is not a multiple or linear combination of row 1 or 2 so therefore, row 1 and 3 are linearly independent. hence, rank = 2

Answer 7

As people have suggested, get it down to reduced form and see how many leading 1's there is :) that is the rank.

Answer 8

So if i get something like

1 4 6
0 3 4
0 0 0

rank is 2?



or

2 5 7
0 0 0
0 0 0


rank is 1?





Or


1 4 5
1 0 0
1 0 4

rank is 3?

Answer 9

no. the third one is incorrect

Answer 10

given an nxm matrix

rankA + nullA = m

Answer 11

third would be 2?

Answer 12

soz third one is correct. it is 3

Answer 13

1 4 5
1 0 0
1 0 4


1 0 0
1 4 5
1 0 4


1 0 0
0 4 5
0 0 4

Answer 14

okay, i think i undertand now



so we reduce to row echelon form and do leading 1's then count the smallest number of rows?

Answer 15

so
1 0 0
0 4 5
0 0 4

becomes

1 0 0
0 1 5/4
0 0 1

and that has three 1's so rank is 3

Answer 16

counting leading ones would suffice. yes, above is correct

Answer 17

thanks all!!